March 2013. GCSE Mathematics (Linear) 1MA0 ... qualifications websites at www
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Education Ltd 2013 ..... C2 (dep M1) for x = 109 with complete reasons. (C1 (dep
M1) for ...
Mark Scheme (Results) March 2013 GCSE Mathematics (Linear) 1MA0 Higher (Non-Calculator) Paper 1H
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March 2013 Publications Code UG035047 All the material in this publication is copyright © Pearson Education Ltd 2013
NOTES ON MARKING PRINCIPLES 1
All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last.
2
Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions.
3
All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme.
4
Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited.
5
Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.
6
Mark schemes will indicate within the table where, and which strands of QWC, are being assessed. The strands are as follows: i) ensure that text is legible and that spelling, punctuation and grammar are accurate so that meaning is clear Comprehension and meaning is clear by using correct notation and labeling conventions. ii) select and use a form and style of writing appropriate to purpose and to complex subject matter Reasoning, explanation or argument is correct and appropriately structured to convey mathematical reasoning. iii) organise information clearly and coherently, using specialist vocabulary when appropriate. The mathematical methods and processes used are coherently and clearly organised and the appropriate mathematical vocabulary used.
7
With working If there is a wrong answer indicated on the answer line always check the working in the body of the script (and on any diagrams), and award any marks appropriate from the mark scheme. If working is crossed out and still legible, then it should be given any appropriate marks, as long as it has not been replaced by alternative work. If it is clear from the working that the “correct” answer has been obtained from incorrect working, award 0 marks. Send the response to review, and discuss each of these situations with your Team Leader. If there is no answer on the answer line then check the working for an obvious answer. Any case of suspected misread loses A (and B) marks on that part, but can gain the M marks. Discuss each of these situations with your Team Leader. If there is a choice of methods shown, then no marks should be awarded, unless the answer on the answer line makes clear the method that has been used.
8
Follow through marks Follow through marks which involve a single stage calculation can be awarded without working since you can check the answer yourself, but if ambiguous do not award. Follow through marks which involve more than one stage of calculation can only be awarded on sight of the relevant working, even if it appears obvious that there is only one way you could get the answer given.
9
Ignoring subsequent work It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is inappropriate for the question: e.g. incorrect cancelling of a fraction that would otherwise be correct It is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect e.g. algebra. Transcription errors occur when candidates present a correct answer in working, and write it incorrectly on the answer line; mark the correct answer.
10
Probability Probability answers must be given a fractions, percentages or decimals. If a candidate gives a decimal equivalent to a probability, this should be written to at least 2 decimal places (unless tenths). Incorrect notation should lose the accuracy marks, but be awarded any implied method marks. If a probability answer is given on the answer line using both incorrect and correct notation, award the marks. If a probability fraction is given then cancelled incorrectly, ignore the incorrectly cancelled answer.
11
Linear equations Full marks can be gained if the solution alone is given on the answer line, or otherwise unambiguously indicated in working (without contradiction elsewhere). Where the correct solution only is shown substituted, but not identified as the solution, the accuracy mark is lost but any method marks can be awarded.
12
Parts of questions Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another.
13
Range of answers Unless otherwise stated, when an answer is given as a range (e.g 3.5 – 4.2) then this is inclusive of the end points (e.g 3.5, 4.2) and includes all numbers within the range (e.g 4, 4.1)
Guidance on the use of codes within this mark scheme M1 – method mark A1 – accuracy mark B1 – Working mark C1 – communication mark QWC – quality of written communication oe – or equivalent cao – correct answer only ft – follow through sc – special case dep – dependent (on a previous mark or conclusion) indep – independent isw – ignore subsequent working
1MA0_1H Question 1
Working 183 × 47 1281 7320 8601 or
Mark
86.01
3
Notes M1 for a complete method to multiply 183 by 47 and attempt at addition (condone one multiplication error) A1 for digits 8601 given as the answer B1 (dep on M1) for correctly writing their answer to 2 decimal places
1
8
3
3
1
4
2 5
8
Answer
7 6
2 2
6 0
×
1
4 7
1
or 100 80 4000 3200 700 560
3 120 21
40 7
4000 + 3200 +120 + 700 + 560 + 21 = 8601 or 183 × 100 = 18 300 183 × 50 = 18 300 ÷ 2 =9150 183 × 3 = 549 9150 – 549 = 8601
1MA0_1H Question 2
3
Working
Answer
Mark
Plot (2, 250) and (3.1, 190)
Plot points
1
B1 for both points plotted accurately
(b)
Relationship
1
B1 for “As the distance from the centre increases the monthly rent decreases” or the nearer you are to the centre the more you have to pay oe (accept negative correlation)
(c)
200 to 260
2
M1 for attempting a correct method, eg a line of best fit or any other indication, on a line that could be used as a line of best fit eg line to graph at x = 2.8or a mark on the line at 2.8 A1 for value in the range 200 to 260
(a)
2 reasons
2
B2 for 2 different reasons from given examples (B1 for 1 reason from given examples)
(a)
Notes
eg No time frame eg No box for less than £10 accept no box for zero or none or £0 eg Overlapping intervals or boxes or £30 and/ or £50 in two boxes (b)
1 reason
1
C1 for reason why the sample is biased eg • they are only in the CD store, • the people in the store are more likely to buy CDs • she needs to ask people outside the store oe
1MA0_1H Question 4
(a)
Working x y
–2 (1)
–1 3
0 (5)
1 7
2 9
(b)
Answer
Mark
Notes
3, 7, 9
2
B2 for all three values correct in the table (B1 for 2 values correct)
graph of y = 2x + 5
2
(From their table of values) M1 ft for plotting at least 2 of their points (any points from their table must be correctly plotted) A1 for correct line from x = –2 to x = +2
1 (Use of y = mx + c) M1 for line drawn with gradient of 2 or line drawn with a y intercept of 5 and a positive gradient) A1 for correct line from x = –2 to x = +2
8 6 4 2 O
2 x
1MA0_1H Question 5
Working
Answer
Mark
Notes
(a)
6n – 3
2
M1 for attempt to establish linear expression in n with coefficient of 6 e.g. 6n + k where k is an integer (accept n = 6n –3 for one mark) A1 cao
(b)
No + Reason
1
C1 ft from their answer to part (a) for decision and explanation eg “ stating no and because all the terms in the sequence are odd and 150 is even” or “no and ‘6n – 3’ = 150, n = 153/6 ... so n is not an integer” or Continuing the sequence to show terms 147 and 153 and state “no as 150 is not in the sequence” oe
1MA0_1H Question 6
Working
Answer
Mark
Notes
(a)
8
1
B1 for 8 (.00)
(b)
550
4
M1 for 600 – 200 ( = 400) M1 for correct method to convert ‘$400’ to £ M1 (dep on the previous M1) for 800 – ‘$400’ in £s A1 for value in the range 540 –560 OR
M1 for correct method to convert $600 and $200 to pounds M1 for ‘375’–‘125’ M1 (dep on the previous M1) 800 –‘250’ A1 for a value in the range 540-560 OR M1 for correct method to convert £800 to dollars M1 for ‘1280’ + 200 – 600 M1 (dep on the previous M1) for attempt to convert ‘$880’ back to £ A1 for value in the range 540 – 560 7
(a)
(b)
7x + 14 = 7 7x = –7
or
x+2=1
6x – 3y
2
M1 for an attempt to combine terms in x or terms in y correctly eg 5x + x(= 6x), 4y – 7y(= – 3y) A1 for 6x – 3y oe
x = –1
2
M1 for correctly expanding the bracket or an attempt to divide both sides by 7 e.g. 7x +14 or x + 2 = 7 ÷ 7 oe A1 cao
1MA0_1H Question 8
Working
Answer
Mark
Notes
09 36
3
M1 for listing 9, 18, 27, 36, 45, ...(at least 3 correct multiples with at most one incorrect) M1 for listing 12, 24, 36, 48, .... (at least 3 correct multiples with at most one incorrect) A1 for 09 36 or 9 36 (am) OR M1 for listing 9.09 9.18 9.27 9.36 ...(at least 3correct times with at most one incorrect) M1 for listing 9.12 9.24 9.36 ... (at least 3 correct times with at most one incorrect) A1 for 09 36 or 9 36 (am) OR M1 for 9 = 3 × 3 or 12 = 2 × 2 × 3 (could be in factor tree) M1 for 9 = 3 × 3 and 12 = 2 × 2 × 3 (could be in a factor tree) A1 for 09 36 or 9 36 (am) SC B2 for 9 36 pm or (after) 36 (minutes) on the answer line
9
(a)
a9
1
B1 for a 4 + 5 or a 9
(b)
9e5f 6
2
B2 cao (B1 for two of 9, e 6 – 1, f 8 – 2 as a product)
(c)
3
1
B1 (accept ± 3 but not just –3)
1MA0_1H Question *10
Working
Answer
Mark
Angle AED = 38 alternate angles are equal Angle ADE = (180 – 38) ÷ 2 = 71 x = 180 – 71 base angles of an isosceles triangle are equal angles in a triangle add up to 180 angles on a straight line sum to 180 OR angle AEF = 142 allied angles/co-interior angles add up to 180 ADE = 142 ÷ 2 = 71 base angles of an isosceles triangle are equal exterior angle of a triangle is equal to the sum of the interior opposite angles , x = 180 – 71 angles in a straight line add to 180 OR Angle AED = 38 alternate angles are equal for angles BAE and AED and BAD and ADC (x) Angle DAE= (180 – 38) ÷ 2 = 71 base angles of an isosceles triangle are equal angles in a triangle add up to 180 Or Angle AED = 38 alternate angles are equal Angle ADE = (180 – 38) ÷ 2 = 71 base angles of an isosceles triangle are equal and angles in a triangle sum to 180 x = 38 + 71 alternate angles BAD and ADC(x) are equal
x = 109
4
Notes B1 for angle AED = 38 or AEF = 142 M1 for a complete method to find one of the base angles of the isosceles triangle C2 (dep M1) for x = 109 with complete reasons (C1 (dep M1) for one reason correctly used and stated)
1MA0_1H Question 11
Working
Answer
Mark
730
5
Notes
5 × 200 ( = 10) oe 100 10 × 350 ( = 35) oe M1 for 100 M1 for
M1 for 6 × ‘10’ or 4 × ‘35’ M1 (dep on M1 earned for a correct method for a percentage calculation) for “60” + “140”+ 530 A1 cao Or M1 for 6 × 200 (= 1200) or 4 × 350 (= 1400)
5 × "1200"(= 60) oe 100 10 × "1400"(= 140) oe M1 for 100 M1 for
M1(dep on M1 earned for a correct method for a percentage calculation) for “60” + “140”+ 530 A1 cao
1MA0_1H Question 12
Working
Answer
Mark
240
4
Notes M1 for 16 × 2 (= 32 girls) M1 for 16 + ‘16 × 2’ (= 48) M1 (dep on the previous M1) for (16 + ‘32’ )× 5 or (16 + ‘32’) × (4 + 1) A1 cao OR M1 for 1 : 2 = 3 parts M1 for 5 schools × 3 parts (= 15 parts) M1 (dep on the previous M1) for ‘15’ parts × 16 A1 cao SC B2 for 176 given on the answer line
13
54
3
M1 for 180 – 360 ÷ 5 or 108 seen as the interior angle of a pentagon M1 (dep on previous M1) for 360 – 2 × ‘108’ – 90 A1 for 54 cao OR M1 for 180 × (5 – 2) (= 540) ÷ 5 or 108 given as the interior angle of a pentagon M1 (dep on previous M1) for 360 – 2 × ‘108’ – 90 A1 for 54 cao
1MA0_1H Question 14
Working
Answer
Mark
Notes
(a)
8, 23, 53, 70, 77, 80
1
B1 cao
(b)
graph
2
M1 ft from their table for at least 5 points plotted correctly at the ends of the intervals provided table values are cumulative, condoning one arithmetic error A1 cao for correct graph with points joined by curve or straight line segments [SC B1 if the shape of the graph is correct and 5 points of their points are not at the ends but consistently within each interval and joined.]
(c)
Readings at 60 and 20 420 to 440 – 280 to 295
(d)
80 – 71 to 74
120 – 160
2
M1 (dep on cf graph) for use of either cf = 20 or cf = 60 A1 ft from a cf graph
6–9
2
M1 (dep on cf graph) for evidence of reading off the cf axis from £530 0n the wages axis (could be the answer) A1ft for 6 - 9
1MA0_1H Question 15
16
17
Working
Answer Required region
Mark 4
Notes M1 arc radius 5 cm centre C M1 bisector of angle BAD M1 line 3 cm from DC A1 for correct region identified (see overlay)
(a)
820 000
1
B1 cao
(b)
3.76 × 10–4
1
B1 cao
(c)
5 × 108
2
M1 for 2.3 ÷ 4.6 × 1012 – 3 oe or 500 000 000 or 0.5 × 109 A1 cao (accept 5.0 × 108
12 13
3
M1 for multiplying throughout by 10 oe or writing LHS as a single fraction e.g 2(4x – 1) + 5(x + 4) = 3× 10 or – 2(4 x − 1) + 5( x + 4) or + 10 M1 (dep) for a complete correct method to obtain linear equation of the form ax = b (condone one arithmetic error in multiplying out the bracket) A1 for 12 oe (decimal equivalent is 0.923…) 13
1MA0_1H Question 18
Working Q at (– 3, 1), (– 6, 1) (–5, 3) (– 3, 3)
Answer
Mark
Notes
Rotation 180° about (–1, 0)
3
M1 for showing R correctly on the grid without showing Q or for showing Q and R correctly on the grid A1 for rotation of 180° A1 for (centre) (–1, 0)
R at (–3, – 1), (–6, – 1), (–5, – 3) (–3, –3)
Or M1 for showing R correctly on the grid without showing Q or for showing Q and R correctly on the grid A1 for Enlargement Scale Factor –1 A1 for centre (–1, 0) NB Award no marks for any correct answer from an incorrect diagram or any Accuracy marks if more than one transformation is given 19
68
3
M1 for angle OBC = 90° or angle OAC = 90° (may be marked on the diagram or used in subsequent working) M1 for correct method to find angle BOC or AOC or AOB e.g. angle BOC = 180 – 90 – 34 (= 56) or angle AOC = 180 – 90 – 34 (=56) or angle AOB = 180 – 2 × 34 (= 112) A1 cao NB (68 must be clearly stated as an answer and not just seen on diagram)
1MA0_1H Question 20 (a)(i)
*21
Working
Answer (x – 9)(x – 3)
Mark 3
Notes M1 for (x ± 9)(x ± 3) A1 for (x – 9)(x – 3)
(ii)
x=9, x=3
(b)
(y + 10)(y –10)
1
B1 for (y + 10)(y –10)
proof
4
M1 for any two consecutive integers expressed algebraically eg n and n +1
(n + 1)2 – n2 = n2 + 2n + 1 – n2 = 2n + 1 (n + 1) + n = 2n + 1
B1 cao
M1(dep on M1) for the difference between the squares of ‘two consecutive integers’ expressed algebraically eg (n + 1)2 – n2
OR (n + 1)2 – n2 = (n + 1 + n)(n + 1 – n) = (2n + 1)(1) = 2n + 1 (n + 1) + n = 2n + 1
A1 for correct expansion and simplification of difference of squares, eg 2n + 1 C1 (dep on M2A1) for showing statement is correct, eg n + n + 1 = 2n + 1 and (n + 1)2 – n2 = 2n + 1 from correct supporting algebra
OR n2 – (n + 1)2 = n2 – (n2 + 2n + 1) = –2n – 1 = – (2n + 1) Difference is 2n + 1 (n + 1) + n = 2n + 1 22
Vertices at (–2, –4), (–4, –4), (–4, –6), (–2, –5)
Correct diagram
3
M1 for a similar shape in the correct orientation in the third quadrant M1 for an image in the correct orientation of the correct size A1 cao
1MA0_1H Question 23
24
Working
EE + CC + HH
Answer 75π
Mark 3
76 110
5
Notes M1 for (4 × π × 52 ) ÷ 2 oe M1 for π × 52 oe A1 for 75π accept 235.5 Condone the use of π = 3.14… M1 for use of 10 as denominator for 2nd probability M1 for 4 × 3 or 5 × 4 or 2 × 1 11 10 11 10 11 10 M1 for 4 × 3 + 5 × 4 + 2 × 1 ⎛⎜ = 34 ⎞⎟ 11 10 11 10 11 10 ⎝ 110 ⎠ M1 (dep on previous M1 for 1 – ‘ 34 ’ 110
A1 for Or EC+EH+CE+CH+HE+HC
76 110
oe
Or M1 for use of 10 as denominator for 2nd probability M1 for 4 × 5 or 4 × 2 or 5 × 4 or 5 × 2 or 2 × 4 or M2 for
11 10 4×5 11 10
+
2× 5 11 10 11 10 11 10 11 10 11 10 4× 2 + 5× 4 + 5× 2 + 2× 4 + 2× 5 11 10 11 10 11 10 11 10 11 10
(M1 for at least 3 of these) A1 for 76 oe 110
Or
E,not E+ C,not C + H,not H
Or M1 for use of 10 as denominator for 2nd probability M1 for 4 × 7 or 5 × 6 or 2 × 9 11 10 11 10 11 10 4 7 5 6 2 9 M2 for × + × + × 11 10 11 10 11 10 (M1 for two of these added) A1 for 76 oe PTO for SC’s 110
1MA0_1H Question
Working
Answer
Mark
Notes SC: B2 for
76 121
SC: B1 for × + × + × (= Or × + × + × + × + × Or × + × + × 25
(a)
sketch
(b)
y = f (x – 6)
) +
×
M1 for inverting the parabola, so maximum is at ( –2, 0 ) A1 for parabola passing through all three of the points ( –2, 0), (0, –4), ( –4, –4) 1
B1 for y = f (x – 6) or y = (x – 4)2 oe
1MA0_1H Question 26 (a) (b)
Working
Answer 6b – 3a
Mark 1 4
Notes B1 for 6b – 3a oe M1 for AX =
1 1 AB or ’(6b – 3a)’ or ft to 2b – a 3 3
M1 for OY = OB + BY = 6b + 5a – b (= 5b + 5a ) oe M1 for OX = 3a + ‘2b – a’ = 2a + 2b oe Or OX = 6b –
‘(6b – 3a)’ (= 2a + 2b) oe
C1 for 2 OY = 2 ×5(a + b) = 2(a + b) = OX 5
5
4. y 10
9 8
7 6
5 4
3 2
1 -2
–1
O
–1
1
2
x
18. y 7 6 5 4 3 Q
P
2 1
-7
-6
-5
-4 R
-3
-2
-1
O 1 -2 -3 -4 -5 -6 -7
1
2
3
4
5
6
7
x
22 y 8
6
4
2
-
-6
-
-
O -
-
-
-
2
4
6
8
x
25(a) y 4
2
-4
-2
0
-2
-4
2
4
x
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